Let $R$ be the region in the first, second, and third quadrants enclosed by the polar curve $r(\theta)=1+2\cos^2(\theta)$ and the coordinate axes. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{0}^{3\pi}\left( \dfrac{1}{4}+\cos^2(\theta)+\cos^4(\theta)\right)d\theta$ (Choice B) B $ \int_{0}^{\scriptsize\dfrac{3\pi}{2}}\left( \dfrac{1}{4}+\cos^2(\theta)+\cos^4(\theta)\right)d\theta$ (Choice C) C $ \int_{0}^{\scriptsize\dfrac{3\pi}{2}}\left( \dfrac{1}{2}+2\cos^2(\theta)+2\cos^4(\theta)\right)d\theta$ (Choice D) D $ \int_{0}^{3\pi}\left( \dfrac{1}{2}+2\cos^2(\theta)+2\cos^4(\theta)\right)d\theta$
Solution: This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ We know $r(\theta)$ but we still need to figure out $\alpha$ and $\beta$. Since $R$ is in the first, second, and third quadrants, and $r(\theta)>0$ for all values of $\theta$, its boundaries are $\alpha=0$ and $\beta=\dfrac{3\pi}{2}$. Let's plug ${r(\theta)=1+2\cos^2(\theta)}$, ${\alpha=0}$, and ${\beta=\dfrac{3\pi}{2}}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{0}}^{{\scriptsize\dfrac{3\pi}{2}}}\dfrac{1}{2}\left({1+2\cos^2(\theta)}\right)^{2}d\theta \\\\ &= \int_{0}^{\scriptsize\dfrac{3\pi}{2}}\dfrac{1}{2}\left( 1+4\cos^2(\theta)+4\cos^4(\theta)\right)d\theta \\\\ &= \int_{0}^{\scriptsize\dfrac{3\pi}{2}}\left( \dfrac{1}{2}+2\cos^2(\theta)+2\cos^4(\theta)\right)d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{0}^{\scriptsize\dfrac{3\pi}{2}}\left( \dfrac{1}{2}+2\cos^2(\theta)+2\cos^4(\theta)\right)d\theta$